If x is a real number and \((\mathrm{x}+5)(\mathrm{x}-3)\) is a negative real number, then the value of which of the...

1. Translate the problem requirements: We need to find when \((\mathrm{x}+5)(\mathrm{x}-3) < 0\), then determine which of the three given expressions must also be negative for all such x values.
2. Determine the constraint on x: Use sign analysis to find the range of x values that make \((\mathrm{x}+5)(\mathrm{x}-3)\) negative.
3. Test each expression systematically: For each of the three expressions, check whether it's always negative, always positive, or sometimes positive/negative within our constraint range.
4. Eliminate based on counterexamples: Use specific test values within the constraint range to quickly eliminate expressions that can be positive.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for. We have the expression \((\mathrm{x}+5)(\mathrm{x}-3)\) and we're told this equals a negative number. Think of this like asking: "When does the product of two things give us a negative result?"

In everyday terms, a product is negative when one factor is positive and the other is negative - just like when you multiply a positive number by a negative number.

So we need \((\mathrm{x}+5)(\mathrm{x}-3) < 0\), and then we want to find which of the three given expressions (I, II, or III) must also be negative for any x value that satisfies our condition.

Process Skill: TRANSLATE - Converting the problem statement into a clear mathematical requirement

2. Determine the constraint on x

Now let's figure out when \((\mathrm{x}+5)(\mathrm{x}-3)\) is negative. Think about this step by step:

The expression \((\mathrm{x}+5)\) equals zero when \(\mathrm{x} = -5\)
The expression \((\mathrm{x}-3)\) equals zero when \(\mathrm{x} = 3\)

These two points divide our number line into three regions:

- When \(\mathrm{x} < -5\): both \((\mathrm{x}+5)\) and \((\mathrm{x}-3)\) are negative, so their product is positive
- When \(-5 < \mathrm{x} < 3\): \((\mathrm{x}+5)\) is positive and \((\mathrm{x}-3)\) is negative, so their product is negative
- When \(\mathrm{x} > 3\): both \((\mathrm{x}+5)\) and \((\mathrm{x}-3)\) are positive, so their product is positive

Therefore, \((\mathrm{x}+5)(\mathrm{x}-3) < 0\) exactly when \(-5 < \mathrm{x} < 3\).

This means x must be between -5 and 3 (not including the endpoints).

Process Skill: VISUALIZE - Using number line analysis to understand when the product is negative

3. Test each expression systematically

Now we'll check each expression to see if it's always negative when \(-5 < \mathrm{x} < 3\).

Expression I: \(3-\mathrm{x}\)
When \(-5 < \mathrm{x} < 3\), let's see what happens to \(3-\mathrm{x}\):

- If x is close to -5, then \(3-\mathrm{x}\) is close to \(3-(-5) = 8\) (positive)
- If x is close to 3, then \(3-\mathrm{x}\) is close to \(3-3 = 0\) (approaches zero from positive side)

So \(3-\mathrm{x}\) is always positive in our range. Expression I is NOT always negative.

Expression II: \(3\mathrm{x} - 7\)
Let's test some values in our range \(-5 < \mathrm{x} < 3\):

- When \(\mathrm{x} = -4\): \(3(-4) - 7 = -12 - 7 = -19\) (negative)
- When \(\mathrm{x} = 0\): \(3(0) - 7 = -7\) (negative)
- When \(\mathrm{x} = 2\): \(3(2) - 7 = 6 - 7 = -1\) (negative)

But let's check if this is always negative. For \(3\mathrm{x} - 7\) to equal zero: \(3\mathrm{x} = 7\), so \(\mathrm{x} = \frac{7}{3} \approx 2.33\)
Since \(\frac{7}{3}\) is between -5 and 3, there are values in our range where \(3\mathrm{x} - 7\) is positive (when \(\mathrm{x} > \frac{7}{3}\)).

So Expression II is NOT always negative.

Expression III: \((\mathrm{x}+7)(\mathrm{x}-3)\)
Notice that this expression has the same factor \((\mathrm{x}-3)\) as our original constraint, but \((\mathrm{x}+5)\) is replaced by \((\mathrm{x}+7)\).

When \(-5 < \mathrm{x} < 3\):

- \((\mathrm{x}-3)\) is always negative (since \(\mathrm{x} < 3\))
- \((\mathrm{x}+7)\) is always positive (since \(\mathrm{x} > -5\), which means \(\mathrm{x}+7 > 2\))

Since \((\mathrm{x}+7)\) is positive and \((\mathrm{x}-3)\) is negative throughout our entire range, their product \((\mathrm{x}+7)(\mathrm{x}-3)\) is always negative.

Process Skill: APPLY CONSTRAINTS - Systematically checking each expression against our constraint \(-5 < \mathrm{x} < 3\)

4. Eliminate based on counterexamples

From our analysis:

- Expression I \((3-\mathrm{x})\): Always positive in our range, so eliminated
- Expression II \((3\mathrm{x}-7)\): Can be positive when \(\mathrm{x} > \frac{7}{3} \approx 2.33\), so eliminated
- Expression III \((\mathrm{x}+7)(\mathrm{x}-3)\): Always negative in our range, so this is our answer

To double-check with a specific value, let's use \(\mathrm{x} = 0\) (which satisfies \(-5 < \mathrm{x} < 3\)):

- Expression I: \(3-0 = 3\) (positive) ✗
- Expression II: \(3(0)-7 = -7\) (negative) ✓
- Expression III: \((0+7)(0-3) = 7(-3) = -21\) (negative) ✓

Now let's try \(\mathrm{x} = 2.5\):

- Expression I: \(3-2.5 = 0.5\) (positive) ✗
- Expression II: \(3(2.5)-7 = 7.5-7 = 0.5\) (positive) ✗
- Expression III: \((2.5+7)(2.5-3) = 9.5(-0.5) = -4.75\) (negative) ✓

Only Expression III is consistently negative.

5. Final Answer

Only Expression III, \((\mathrm{x}+7)(\mathrm{x}-3)\), must be negative when \((\mathrm{x}+5)(\mathrm{x}-3)\) is negative.

The correct answer is C. III only.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "negative real number" requirement
Students may focus only on finding when each expression I, II, and III is negative, without first establishing the constraint that \((\mathrm{x}+5)(\mathrm{x}-3) < 0\). They skip the crucial step of determining that x must be between -5 and 3, leading them to test expressions across all real numbers instead of within the specific range.

2. Confusing "must be negative" with "can be negative"
Students often misread the question as asking which expressions "can be negative" rather than which "must be negative." This leads them to select expressions that are sometimes negative within the range, rather than identifying expressions that are always negative when the constraint is satisfied.

3. Incomplete sign analysis approach
Some students attempt to solve this algebraically by expanding expressions or using complex methods, rather than recognizing this as a sign analysis problem that requires examining when products of linear factors change signs at their zeros.

Errors while executing the approach

1. Incorrect interval determination for the constraint
When finding where \((\mathrm{x}+5)(\mathrm{x}-3) < 0\), students frequently make sign errors. They may incorrectly conclude that the product is negative when \(\mathrm{x} < -5\) or \(\mathrm{x} > 3\), missing that the product is actually negative only between the zeros: \(-5 < \mathrm{x} < 3\).

2. Inadequate testing of boundary behavior
Students often test only one or two values (like \(\mathrm{x} = 0\)) within the valid range, missing cases where expressions change sign within the interval. For Expression II \((3\mathrm{x}-7)\), they might miss that it becomes positive when \(\mathrm{x} > \frac{7}{3} \approx 2.33\), which is still within the constraint range.

3. Sign analysis errors for individual expressions
When evaluating Expression III: \((\mathrm{x}+7)(\mathrm{x}-3)\), students may incorrectly determine the sign of \((\mathrm{x}+7)\) when \(-5 < \mathrm{x} < 3\), forgetting that since \(\mathrm{x} > -5\), we have \(\mathrm{x}+7 > 2\), making \((\mathrm{x}+7)\) always positive in this range.

Errors while selecting the answer

1. Selecting based on partial analysis
After finding that both Expressions II and III are negative for some test values (like \(\mathrm{x} = 0\)), students may hastily select "E. II and III only" without verifying that these expressions remain negative throughout the entire valid range \(-5 < \mathrm{x} < 3\).

2. Misreading Roman numeral format
Students may correctly identify that only Expression III must be negative, but then incorrectly match this to answer choice "A. I only" instead of "C. III only," confusing the Roman numerals with the answer choice letters.

Alternate Solutions

Smart Numbers Approach

Step 1: Find the constraint range
For \((\mathrm{x}+5)(\mathrm{x}-3) < 0\), we need one factor positive and one negative.
• \(\mathrm{x}+5 = 0\) when \(\mathrm{x} = -5\)
• \(\mathrm{x}-3 = 0\) when \(\mathrm{x} = 3\)
• The product is negative when \(-5 < \mathrm{x} < 3\)

Step 2: Choose strategic test values
Let's use smart numbers within our constraint range \(-5 < \mathrm{x} < 3\):
• \(\mathrm{x} = 0\) (convenient for calculations, middle of range)
• \(\mathrm{x} = -2\) (negative value in range)
• \(\mathrm{x} = 2\) (positive value in range)

Step 3: Test each expression with our smart numbers

For Expression I: \(3-\mathrm{x}\)
• When \(\mathrm{x} = 0\): \(3-0 = 3\) (positive)
• When \(\mathrm{x} = -2\): \(3-(-2) = 5\) (positive)
• When \(\mathrm{x} = 2\): \(3-2 = 1\) (positive)
Expression I is always positive, so it's eliminated.

For Expression II: \(3\mathrm{x}-7\)
• When \(\mathrm{x} = 0\): \(3(0)-7 = -7\) (negative)
• When \(\mathrm{x} = -2\): \(3(-2)-7 = -13\) (negative)
• When \(\mathrm{x} = 2\): \(3(2)-7 = -1\) (negative)
All tests are negative, but let's check a value closer to 3:
• When \(\mathrm{x} = 2.5\): \(3(2.5)-7 = 0.5\) (positive)
Expression II can be positive, so it's eliminated.

For Expression III: \((\mathrm{x}+7)(\mathrm{x}-3)\)
• When \(\mathrm{x} = 0\): \((0+7)(0-3) = 7(-3) = -21\) (negative)
• When \(\mathrm{x} = -2\): \((-2+7)(-2-3) = 5(-5) = -25\) (negative)
• When \(\mathrm{x} = 2\): \((2+7)(2-3) = 9(-1) = -9\) (negative)
Since \(\mathrm{x}+7 > 0\) for all x in our range (-5,3) and \(\mathrm{x}-3 < 0\) for all x in our range, Expression III is always negative.

Answer: C (III only)